Mixing
Sobolev integrability implies weak flatness?
$begingroup$
Let $B_1subset mathbb R^n$ be the unit ball, and suppose $uin W^{1,2}(B_1)$, i.e. $int_{B_1} u^2+|Du|^2,dxle K<infty$. Moreover, suppose $u$ is nonnegative, with $essinf u=0$.
Let us also define the distribution function $rho(t)=|{ule t}|$. The nonnegativity and infimum condition simply mean $rho(t)=0$ precisely when $t< 0$.
Question 1 (answered by supinf): can we deduce from the above that $rho(t)ge C t^alpha$ for some $C,alpha>0$ depending on $K,n$?
Question 2 (supinf): Replace $rho(t)ge Ct^alpha$ with $rho(t)ge rho(0)+Ct^alpha$? If $rho(0)=0$?
Question 3: Can we deduce $rho(t)ge Ct^{alpha}$ for small $tge 0$, and some $C,alpha>0$ depending on $K$?
Here, $rho(0)neq 0$ is allowed, and this smallness should remove counterexamples related to the putative ``power growth" of $rho(t)$.
Heuristics:
If $u$ is smooth, then this is certainly true, with $alpha=n$. So it is reasonable that $int |Du|^2<infty$ would have a similar effect.
If $n=1$, one would initially guess that the family $u_k(x)=|x|^{1/k}$ would comprise a counterexample, since the distribution function decays with $k$, but in fact, this family is not in $W^{1,2}$.
Note that the distribution function also vanishes for the $W^{1,2}$ family $u_k(x)=U(|x|)^{1/k}$, where $U(x)=x-xln x$. But it can be checked that the Poincare constant $int |Du|^2/int u^2$ blows up, so $K=K(k)$ is unbounded.
If $u$ is Lipschitz, the distribution function is somewhat related to the $W^{1,2}$ norm using the co-area formula:
$$
int_{B_1} |Du|=int_{0}^infty |{u=t}|dt.
$$
real-analysis pde sobolev-spaces
asked Jan 31 at 1:36
user254433user254433
2,541712
$endgroup$
add a comment |
$begingroup$
Let $B_1subset mathbb R^n$ be the unit ball, and suppose $uin W^{1,2}(B_1)$, i.e. $int_{B_1} u^2+|Du|^2,dxle K<infty$. Moreover, suppose $u$ is nonnegative, with $essinf u=0$.
Let us also define the distribution function $rho(t)=|{ule t}|$. The nonnegativity and infimum condition simply mean $rho(t)=0$ precisely when $t< 0$.
Question 1 (answered by supinf): can we deduce from the above that $rho(t)ge C t^alpha$ for some $C,alpha>0$ depending on $K,n$?
Question 2 (supinf): Replace $rho(t)ge Ct^alpha$ with $rho(t)ge rho(0)+Ct^alpha$? If $rho(0)=0$?
Question 3: Can we deduce $rho(t)ge Ct^{alpha}$ for small $tge 0$, and some $C,alpha>0$ depending on $K$?
Here, $rho(0)neq 0$ is allowed, and this smallness should remove counterexamples related to the putative ``power growth" of $rho(t)$.
Heuristics:
If $u$ is smooth, then this is certainly true, with $alpha=n$. So it is reasonable that $int |Du|^2<infty$ would have a similar effect.
If $n=1$, one would initially guess that the family $u_k(x)=|x|^{1/k}$ would comprise a counterexample, since the distribution function decays with $k$, but in fact, this family is not in $W^{1,2}$.
Note that the distribution function also vanishes for the $W^{1,2}$ family $u_k(x)=U(|x|)^{1/k}$, where $U(x)=x-xln x$. But it can be checked that the Poincare constant $int |Du|^2/int u^2$ blows up, so $K=K(k)$ is unbounded.
If $u$ is Lipschitz, the distribution function is somewhat related to the $W^{1,2}$ norm using the co-area formula:
$$
int_{B_1} |Du|=int_{0}^infty |{u=t}|dt.
$$
real-analysis pde sobolev-spaces
asked Jan 31 at 1:36
user254433user254433
2,541712
$endgroup$
1
$begingroup$
Should $rho(t)geq C t^alpha$ only hold for nonnegative $t$?
$endgroup$
– supinf
Jan 31 at 6:51
add a comment |
$begingroup$
Let $B_1subset mathbb R^n$ be the unit ball, and suppose $uin W^{1,2}(B_1)$, i.e. $int_{B_1} u^2+|Du|^2,dxle K<infty$. Moreover, suppose $u$ is nonnegative, with $essinf u=0$.
Let us also define the distribution function $rho(t)=|{ule t}|$. The nonnegativity and infimum condition simply mean $rho(t)=0$ precisely when $t< 0$.
Question 1 (answered by supinf): can we deduce from the above that $rho(t)ge C t^alpha$ for some $C,alpha>0$ depending on $K,n$?
Question 2 (supinf): Replace $rho(t)ge Ct^alpha$ with $rho(t)ge rho(0)+Ct^alpha$? If $rho(0)=0$?
Question 3: Can we deduce $rho(t)ge Ct^{alpha}$ for small $tge 0$, and some $C,alpha>0$ depending on $K$?
Here, $rho(0)neq 0$ is allowed, and this smallness should remove counterexamples related to the putative ``power growth" of $rho(t)$.
Heuristics:
If $u$ is smooth, then this is certainly true, with $alpha=n$. So it is reasonable that $int |Du|^2<infty$ would have a similar effect.
If $n=1$, one would initially guess that the family $u_k(x)=|x|^{1/k}$ would comprise a counterexample, since the distribution function decays with $k$, but in fact, this family is not in $W^{1,2}$.
Note that the distribution function also vanishes for the $W^{1,2}$ family $u_k(x)=U(|x|)^{1/k}$, where $U(x)=x-xln x$. But it can be checked that the Poincare constant $int |Du|^2/int u^2$ blows up, so $K=K(k)$ is unbounded.
If $u$ is Lipschitz, the distribution function is somewhat related to the $W^{1,2}$ norm using the co-area formula:
$$
int_{B_1} |Du|=int_{0}^infty |{u=t}|dt.
$$
real-analysis pde sobolev-spaces
asked Jan 31 at 1:36
user254433user254433
2,541712
$endgroup$
Let $B_1subset mathbb R^n$ be the unit ball, and suppose $uin W^{1,2}(B_1)$, i.e. $int_{B_1} u^2+|Du|^2,dxle K<infty$. Moreover, suppose $u$ is nonnegative, with $essinf u=0$.
Let us also define the distribution function $rho(t)=|{ule t}|$. The nonnegativity and infimum condition simply mean $rho(t)=0$ precisely when $t< 0$.
Question 1 (answered by supinf): can we deduce from the above that $rho(t)ge C t^alpha$ for some $C,alpha>0$ depending on $K,n$?
Question 2 (supinf): Replace $rho(t)ge Ct^alpha$ with $rho(t)ge rho(0)+Ct^alpha$? If $rho(0)=0$?
Question 3: Can we deduce $rho(t)ge Ct^{alpha}$ for small $tge 0$, and some $C,alpha>0$ depending on $K$?
Here, $rho(0)neq 0$ is allowed, and this smallness should remove counterexamples related to the putative ``power growth" of $rho(t)$.
Heuristics:
If $u$ is smooth, then this is certainly true, with $alpha=n$. So it is reasonable that $int |Du|^2<infty$ would have a similar effect.
If $n=1$, one would initially guess that the family $u_k(x)=|x|^{1/k}$ would comprise a counterexample, since the distribution function decays with $k$, but in fact, this family is not in $W^{1,2}$.
Note that the distribution function also vanishes for the $W^{1,2}$ family $u_k(x)=U(|x|)^{1/k}$, where $U(x)=x-xln x$. But it can be checked that the Poincare constant $int |Du|^2/int u^2$ blows up, so $K=K(k)$ is unbounded.
If $u$ is Lipschitz, the distribution function is somewhat related to the $W^{1,2}$ norm using the co-area formula:
$$
int_{B_1} |Du|=int_{0}^infty |{u=t}|dt.
$$
real-analysis pde sobolev-spaces
real-analysis pde sobolev-spaces
asked Jan 31 at 1:36
user254433user254433
2,541712
asked Jan 31 at 1:36
user254433user254433
2,541712
edited Jan 31 at 8:08
user254433
asked Jan 31 at 1:36
user254433user254433
2,541712
asked Jan 31 at 1:36
user254433user254433
2,541712
asked Jan 31 at 1:36
user254433user254433
2,541712
2,541712
1
$begingroup$
Should $rho(t)geq C t^alpha$ only hold for nonnegative $t$?
$endgroup$
– supinf
Jan 31 at 6:51
add a comment |
1
$begingroup$
Should $rho(t)geq C t^alpha$ only hold for nonnegative $t$?
$endgroup$
– supinf
Jan 31 at 6:51
1
1
$begingroup$
Should $rho(t)geq C t^alpha$ only hold for nonnegative $t$?
$endgroup$
– supinf
Jan 31 at 6:51
$begingroup$
Should $rho(t)geq C t^alpha$ only hold for nonnegative $t$?
$endgroup$
– supinf
Jan 31 at 6:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No this is not true.
Consider the example $u=0$.
Then $rho(t) = |B_1|$ if $tgeq 0$ according to the definition of $rho$.
Then for $C,alpha>0$ we have
$$
lim_{ttoinfty}rho(t)= | B_1 | < infty = lim_{ttoinfty} C t^alpha.
$$
Thus
$rho(t)geq Ct^alpha$ cannot hold for every $tgeq 0$.
question 2 and general remarks:
For question 2, the function $u=1$ is a similar counterexample (here we have $rho(0)=0$).
Fundamentally, the problem is that
$$
rho(t) leq |B_1| < infty
qquad forall tinmathbb R
$$
is true for all functions $u in W^{1,2}(B_1)$,
whereas
$$
lim_{ttoinfty} C t^alpha = infty
$$
holds for all $C>0,alpha>0$.
So slight modifications will probably not save your conjecture.
answered Jan 31 at 6:54
supinfsupinf
6,7561129
$endgroup$
$begingroup$
for $rho(t)geq rho(0)+C t^alpha$ the function $u=1$ could be a counterexample. I edited.
$endgroup$
– supinf
Jan 31 at 8:00
$begingroup$
Thank you! I thought about it more, and I think adding a small $t$ condition is the best way to eliminate "flat" counterexamples.
$endgroup$
– user254433
Jan 31 at 8:02
add a comment |
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$begingroup$
No this is not true.
Consider the example $u=0$.
Then $rho(t) = |B_1|$ if $tgeq 0$ according to the definition of $rho$.
Then for $C,alpha>0$ we have
$$
lim_{ttoinfty}rho(t)= | B_1 | < infty = lim_{ttoinfty} C t^alpha.
$$
Thus
$rho(t)geq Ct^alpha$ cannot hold for every $tgeq 0$.
question 2 and general remarks:
For question 2, the function $u=1$ is a similar counterexample (here we have $rho(0)=0$).
Fundamentally, the problem is that
$$
rho(t) leq |B_1| < infty
qquad forall tinmathbb R
$$
is true for all functions $u in W^{1,2}(B_1)$,
whereas
$$
lim_{ttoinfty} C t^alpha = infty
$$
holds for all $C>0,alpha>0$.
So slight modifications will probably not save your conjecture.
answered Jan 31 at 6:54
supinfsupinf
6,7561129
$endgroup$
$begingroup$
for $rho(t)geq rho(0)+C t^alpha$ the function $u=1$ could be a counterexample. I edited.
$endgroup$
– supinf
Jan 31 at 8:00
$begingroup$
Thank you! I thought about it more, and I think adding a small $t$ condition is the best way to eliminate "flat" counterexamples.
$endgroup$
– user254433
Jan 31 at 8:02
add a comment |
$begingroup$
No this is not true.
Consider the example $u=0$.
Then $rho(t) = |B_1|$ if $tgeq 0$ according to the definition of $rho$.
Then for $C,alpha>0$ we have
$$
lim_{ttoinfty}rho(t)= | B_1 | < infty = lim_{ttoinfty} C t^alpha.
$$
Thus
$rho(t)geq Ct^alpha$ cannot hold for every $tgeq 0$.
question 2 and general remarks:
For question 2, the function $u=1$ is a similar counterexample (here we have $rho(0)=0$).
Fundamentally, the problem is that
$$
rho(t) leq |B_1| < infty
qquad forall tinmathbb R
$$
is true for all functions $u in W^{1,2}(B_1)$,
whereas
$$
lim_{ttoinfty} C t^alpha = infty
$$
holds for all $C>0,alpha>0$.
So slight modifications will probably not save your conjecture.
answered Jan 31 at 6:54
supinfsupinf
6,7561129
$endgroup$
$begingroup$
for $rho(t)geq rho(0)+C t^alpha$ the function $u=1$ could be a counterexample. I edited.
$endgroup$
– supinf
Jan 31 at 8:00
$begingroup$
Thank you! I thought about it more, and I think adding a small $t$ condition is the best way to eliminate "flat" counterexamples.
$endgroup$
– user254433
Jan 31 at 8:02
add a comment |
$begingroup$
No this is not true.
Consider the example $u=0$.
Then $rho(t) = |B_1|$ if $tgeq 0$ according to the definition of $rho$.
Then for $C,alpha>0$ we have
$$
lim_{ttoinfty}rho(t)= | B_1 | < infty = lim_{ttoinfty} C t^alpha.
$$
Thus
$rho(t)geq Ct^alpha$ cannot hold for every $tgeq 0$.
question 2 and general remarks:
For question 2, the function $u=1$ is a similar counterexample (here we have $rho(0)=0$).
Fundamentally, the problem is that
$$
rho(t) leq |B_1| < infty
qquad forall tinmathbb R
$$
is true for all functions $u in W^{1,2}(B_1)$,
whereas
$$
lim_{ttoinfty} C t^alpha = infty
$$
holds for all $C>0,alpha>0$.
So slight modifications will probably not save your conjecture.
answered Jan 31 at 6:54
supinfsupinf
6,7561129
$endgroup$
No this is not true.
Consider the example $u=0$.
Then $rho(t) = |B_1|$ if $tgeq 0$ according to the definition of $rho$.
Then for $C,alpha>0$ we have
$$
lim_{ttoinfty}rho(t)= | B_1 | < infty = lim_{ttoinfty} C t^alpha.
$$
Thus
$rho(t)geq Ct^alpha$ cannot hold for every $tgeq 0$.
question 2 and general remarks:
For question 2, the function $u=1$ is a similar counterexample (here we have $rho(0)=0$).
Fundamentally, the problem is that
$$
rho(t) leq |B_1| < infty
qquad forall tinmathbb R
$$
is true for all functions $u in W^{1,2}(B_1)$,
whereas
$$
lim_{ttoinfty} C t^alpha = infty
$$
holds for all $C>0,alpha>0$.
So slight modifications will probably not save your conjecture.
answered Jan 31 at 6:54
supinfsupinf
6,7561129
edited Jan 31 at 8:00
answered Jan 31 at 6:54
supinfsupinf
6,7561129
answered Jan 31 at 6:54
supinfsupinf
6,7561129
answered Jan 31 at 6:54
supinfsupinf
6,7561129
6,7561129
$begingroup$
for $rho(t)geq rho(0)+C t^alpha$ the function $u=1$ could be a counterexample. I edited.
$endgroup$
– supinf
Jan 31 at 8:00
$begingroup$
Thank you! I thought about it more, and I think adding a small $t$ condition is the best way to eliminate "flat" counterexamples.
$endgroup$
– user254433
Jan 31 at 8:02
add a comment |
$begingroup$
for $rho(t)geq rho(0)+C t^alpha$ the function $u=1$ could be a counterexample. I edited.
$endgroup$
– supinf
Jan 31 at 8:00
$begingroup$
Thank you! I thought about it more, and I think adding a small $t$ condition is the best way to eliminate "flat" counterexamples.
$endgroup$
– user254433
Jan 31 at 8:02
$begingroup$
for $rho(t)geq rho(0)+C t^alpha$ the function $u=1$ could be a counterexample. I edited.
$endgroup$
– supinf
Jan 31 at 8:00
$begingroup$
for $rho(t)geq rho(0)+C t^alpha$ the function $u=1$ could be a counterexample. I edited.
$endgroup$
– supinf
Jan 31 at 8:00
$begingroup$
Thank you! I thought about it more, and I think adding a small $t$ condition is the best way to eliminate "flat" counterexamples.
$endgroup$
– user254433
Jan 31 at 8:02
$begingroup$
Thank you! I thought about it more, and I think adding a small $t$ condition is the best way to eliminate "flat" counterexamples.
$endgroup$
– user254433
Jan 31 at 8:02
add a comment |
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$begingroup$
Should $rho(t)geq C t^alpha$ only hold for nonnegative $t$?
$endgroup$
– supinf
Jan 31 at 6:51